Question

Tangent lines Find an equation of the line tangent to the graph of \(f\) at the given point. $$f(x)=\sin ^{-1} \frac{x}{4} ;(2, \pi / 6)$$

Short answer

Answer: The equation of the tangent line is \(y - \frac{\pi}{6} = \frac{1}{2\sqrt{3}}(x - 2)\).

Step-by-step solution

Find the derivative of the function f(x)

To find the slope of the tangent line at the given point, we need to find the derivative of the function \(f(x)\). The function is given in the form of inverse trigonometric function: $$f(x) = \sin^{-1}\frac{x}{4}$$ First, let \(y = \sin^{-1}\frac{x}{4}\). This means \(\sin y = \frac{x}{4}\). Now, we need to differentiate both sides of the equation with respect to x. We get: $$\cos y \frac{dy}{dx} = \frac{1}{4}$$ We know \(y = \pi/6\) and \(x = 2\). Since \(\cos(\pi/6) = \frac{\sqrt{3}}{2}\), we can solve for \(\frac{dy}{dx}\): $$\frac{\sqrt{3}}{2} \frac{dy}{dx} = \frac{1}{4}$$ Now, solve for \(\frac{dy}{dx}\): $$\frac{dy}{dx} = \frac{1}{2\sqrt{3}}$$ This is the slope of the tangent line at the given point.

Use point-slope form to find the equation of the tangent line

The point-slope form of a line is given by \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the given point on the line and \(m\) is the slope. In our case, \(x_1 = 2\), \(y_1 = \pi/6\), and the slope \(m = \frac{1}{2\sqrt{3}}\). Now, substitute these values into the point-slope form: $$y - \frac{\pi}{6} = \frac{1}{2\sqrt{3}}(x - 2)$$ This equation represents the tangent line to the graph of the function \(f(x)\) at the point \((2, \pi/6)\).

Key concepts

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to find an angle when the value of the trigonometric function is known. This process is the reverse of finding the sine, cosine, or tangent of an angle, which makes these functions essential in calculus.
For instance, if we have \( f(x) = \sin^{-1}(x) \), this function will give us the angle whose sine is \( x \). However, since the sine function does not have a unique inverse, inverse trigonometric functions are defined for specific ranges to ensure they are functions. For the \(\sin^{-1}(x)\) function, the range is restricted from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \).

Application in Calculus

To use inverse trigonometric functions in calculus, one must be familiar with their derivatives, as these are vital when solving problems involving rates of change or curvature of graphs. For example, in our exercise \( f(x) = \sin^{-1} \frac{x}{4} \) at the point \( (2, \frac{\pi}{6}) \), understanding the inverse function concept allows us to grasp the relationship between an angle and its corresponding sine value.

Derivative of Trigonometric Functions

The derivative of a function at a point measures the instant rate of change of the function at that point. For trigonometric functions, standard rules apply, such as \( \frac{d}{dx}(\sin x) = \cos x \) and \( \frac{d}{dx}(\cos x) = -\sin x \).
When dealing with inverse trigonometric functions, the derivatives become slightly more complex. For example, the derivative of \( \sin^{-1}(x) \) is \( \frac{1}{\sqrt{1 - x^{2}}} \). This knowledge is crucial when finding the slope of tangent lines to the curves defined by these functions, which is exactly what we needed to do in our exercise.

Working with Inverse Trigonometric Functions

To differentiate an inverse trigonometric function, we must use implicit differentiation. In our case, we started with \( y = \sin^{-1}(\frac{x}{4}) \) and implicitly differentiate both sides with respect to \( x \) to find the derivative \( \frac{dy}{dx} \). Once we obtained \( \(cos y \) = \frac{\sqrt{3}}{2} \), we used the given point to solve for \( y \) and find the resulting slope \( m = \frac{1}{2\sqrt{3}} \) for our tangent line.

Point-Slope Form of a Line

The point-slope form is a straightforward method to write the equation of a line when you know a single point through which the line passes \( (x_1, y_1) \) and the slope \( m \) of the line. The formula is:
\[ y - y_1 = m(x - x_1) \]
This form is especially useful in calculus when finding the equation of a tangent line, as we can often derive the slope of the line from the function's derivative at a particular point.

Finding the Tangent Line

In our exercise's context, the point-slope form is utilized after computing the derivative. Once we know the slope \( m = \frac{dy}{dx} \) at the point of tangency and the coordinates of this point, plugging these values into the point-slope form yields the tangent line's equation, as seen with \( x_1 = 2 \) and \( y_1 = \frac{\pi}{6} \).

Equation of a Tangent Line

The equation of a tangent line to a curve at a given point is found by first determining the slope of the tangent line, which is the value of the derivative of the function at that point. The tangent line is a straight line that just 'touches' the curve at that point and has the same slope as the curve at that point.
In essence, a tangent line represents the instantaneous direction of the curve. Finding the equation requires two steps: identifying the slope and then applying the point-slope form. From our initial derivative work, we found the slope \( m \) and using the point-slope form, we substituted the known coordinates and \( m \) to get the final tangent line equation:
\[ y - \frac{\pi}{6} = \frac{1}{2\sqrt{3}}(x - 2) \]
This line provides the best linear approximation to the curve at the point \( (2, \frac{\pi}{6}) \) and is a specific application of the broader concept within calculus.